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To: Paul Engel who wrote (49958)	8/5/2001 4:48:57 PM
From: ptanner	Read Replies (1) \| Respond to of 275872

Paul, Re: "Do the math - 50% reduction is 175 die/wafer and 2500/175 us PRECISLEY $14+ - his EXACT NUMBER." Yes, $2500/175 is $14. But Dan3 wrote "cuts the yields down to 50%" as you quoted in your earlier message which is not the same as reducing yields by 50%. Dan3's other number for working die cost is $2000/250 is $8. I do not see that Dan3 ever assumed 350 working die per wafer in his figures ($2,000/350 is $5.74). -PT

To: Paul Engel who wrote (49958)	8/5/2001 5:36:21 PM
From: Dan3	Read Replies (1) \| Respond to of 275872

Re: This is what he said: AMD can get about 350 80mm die off an 8" wafer That's the starting point - note that I said die, not good die. Let's work through the whole thing. It's all simple approximations, but easy to follow. ((3.14159 * 100 * 100)/80) - (((3.14159 * 200)/9)/2) = 357 A 200mm wafer has an area of 3.14159 * 100 * 100 (Pi * radius squared) or 31,459 square millimeters (mm2). Divide by the estimated area of .13 AThlon 4 (80 mm2) and we get 393 die. But the perimeter of the wafer is going to intersect a number of die about equal to one side of a die (the square root of 80 which we can approximate with 9). The perimeter is equal to the diameter times Pi (3.14159 * 200) which we then divide by 9 to get 69.8. But, on average, we only lose a half a die for each die on the edge (sometime most of a die, other times just a corner of a die is bad) so we divide that 69.8 by 2 to get 35 die from edge loss. 393 - 35 is 358, but rounding takes it down to 357. There are a number of other factors that takes the number of good die per wafer down from there, but 357-358 is where we start for 80mm2 die from a 200mm wafer.